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Mathematics LibreTexts

Section 8.1 Answers

  • Page ID
    29604
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    1. 

    1. \(\frac{1}{s^{2}}\)
    2. \(\frac{1}{(s+1)^{2}}\)
    3. \(\frac{b}{s^{2}-b^{2}}\)
    4. \(\frac{-2s+5}{(s-1)(s-2)}\)
    5. \(\frac{2}{s^{3}}\)

    2. 

    1. \(\frac{s^{2}+2}{\left[(s-1)^{2}+1\right]\left[(s+1)^{2}+1\right]}\)
    2. \(\frac{2}{s(s^{2}+4)}\)
    3. \(\frac{s^{2}+8}{s(s^{2}+16)}\)
    4. \(\frac{s^{2}-2}{s(s^{2}-4)}\)
    5. \(\frac{4s}{(s^{2}-4)^{2}}\)
    6. \(\frac{1}{s^{2}+4}\)
    7. \(\frac{1}{\sqrt{2}}\frac{s+1}{s^{2}+1}\)
    8. \(\frac{5s}{(s^{2}+4)(s^{2}+9)}\)
    9. \(\frac{s^{3}+2s^{2}+4s+32}{(s^{2}+4)(s^{2}+16)}\)

    4. 

    1. \(f(3-)=-1,\: f(3)=f(3+)=1\)
    2. \(f(1-)=3,\: f(1)=4,\: f(1+)=1\)
    3. \(f\left(\frac{\pi }{2}-\right) =1,\: f\left(\frac{\pi }{2} \right) = f\left(\frac{\pi }{2}+ \right)=2,\: f(\pi -)=0,\: f(\pi )=f(\pi +)=-1\)
    4. \(f(1−) = 1,\: f(1) = 2,\: f(1+) = 1,\: f(2−) = 0,\: f(2) = 3,\: f(2+) = 6\)

    5. 

    1. \(\frac{1-e^{-(s+1)}}{s+1}+\frac{e^{-(s+2)}}{s+2}\)
    2. \(\frac{1}{s}+e^{-4s}\left(\frac{1}{s^{2}}+\frac{3}{s}\right)\)
    3. \(\frac{1-e^{-s}}{s^{2}}\)
    4. \(\frac{1-e^{-(s-1)}}{(s-1)^{2}}\)

    7. \(\mathcal{L}(e^{\lambda t}\cos\omega t)=\frac{(s-\lambda )^{2}-\omega^{2}}{((s-\lambda )^{2}+\omega^{2})^{2}}\quad\mathcal{L}(e^{\lambda t}\sin\omega t)=\frac{2\omega (s-\lambda )}{((s-\lambda )^{2}+\omega ^{2})^{2}} \)

    15. 

    1. \(\tan ^{-1}\frac{\omega }{s},\quad s>0\)
    2. \(\frac{1}{2}\ln\frac{s^{2}}{s^{2}+\omega ^{2}},\quad s>0\)
    3. \(\ln\frac{s-b}{s-a},\quad s>\text{max}(a,b)\)
    4. \(\frac{1}{2}\ln\frac{s^{2}}{s^{2}-1},\quad s>1\)
    5. \(\frac{1}{4}\ln\frac{s^{2}}{s^{2}-4},\quad s>2\)

    18. 

    1. \(\frac{1}{s^{2}}\tanh\frac{s}{2}\)
    2. \(\frac{1}{s}\tanh\frac{s}{4}\)
    3. \(\frac{1}{s^{2}+1}\coth\frac{\pi s}{2}\)
    4. \(\frac{1}{(s^{2}+1)(1-e^{-\pi s})}\)